Minimum Time to Make Rope Colorful

Tested Python solution for LeetCode 1578 with 20 pytest cases. Generate a practice environment with lcpy.

LeetCode 1578, Medium. Topics: Array, String, Dynamic Programming, Greedy. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 20 parametrized pytest cases, and a playground notebook:

lcpy gen -n 1578   # by problem number
lcpy gen -s minimum_time_to_make_rope_colorful   # by problem name

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Problem

Alice has n balloons arranged on a rope. You are given a 0-indexed string colors where colors[i] is the color of the i<sup>th</sup> balloon.

Alice wants the rope to be colorful. She does not want two consecutive balloons to be of the same color, so she asks Bob for help. Bob can remove some balloons from the rope to make it colorful. You are given a 0-indexed integer array neededTime where neededTime[i] is the time (in seconds) that Bob needs to remove the i<sup>th</sup> balloon from the rope.

Return the minimum time Bob needs to make the rope colorful.

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Examples

Input: colors = "abaac", neededTime = [1,2,3,4,5]
Output: 3

Explanation: In the above image, ‘a’ is blue, ‘b’ is red, and ‘c’ is green. Bob can remove the blue balloon at index 2. This takes 3 seconds. There are no longer two consecutive balloons of the same color. Total time = 3.

Input: colors = "abc", neededTime = [1,2,3]
Output: 0

Explanation: The rope is already colorful. Bob does not need to remove any balloons from the rope.

Input: colors = "aabaa", neededTime = [1,2,3,4,1]
Output: 2

Explanation: Bob will remove the balloons at indices 0 and 4. Each balloon takes 1 second to remove. There are no longer two consecutive balloons of the same color. Total time = 1 + 1 = 2.

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Constraints

• n == colors.length == neededTime.length
• 1 <= n <= 10^5
• 1 <= neededTime[i] <= 10^4
• colors consists of lowercase English letters.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(n)
    # Space: O(1)
    def min_cost(self, colors: str, needed_time: list[int]) -> int:
        total = 0
        run_max = needed_time[0]
        for i in range(1, len(colors)):
            if colors[i] == colors[i - 1]:
                if needed_time[i] < run_max:
                    total += needed_time[i]
                else:
                    total += run_max
                    run_max = needed_time[i]
            else:
                run_max = needed_time[i]
        return total

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Complexity

Time	Space
O(n)	O(1)

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Tags

NeetCode All.